Quy đồng 5 trên 3x^3-12x và 3 trên (2x+4).(x+3) 26/08/2021 Bởi Harper Quy đồng 5 trên 3x^3-12x và 3 trên (2x+4).(x+3)
Đáp án: $\begin{array}{l} + )\frac{5}{{3{x^3} – 12x}}\\ = \frac{5}{{3x\left( {{x^2} – 4} \right)}}\\ = \frac{{5.2.\left( {x + 3} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\ = \frac{{10x + 30}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\ + )\frac{3}{{\left( {2x + 4} \right)\left( {x + 3} \right)}}\\ = \frac{3}{{2\left( {x + 2} \right)\left( {x + 3} \right)}}\\ = \frac{{3.3x\left( {x – 2} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\ = \frac{{9{x^2} – 18x}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
+ )\frac{5}{{3{x^3} – 12x}}\\
= \frac{5}{{3x\left( {{x^2} – 4} \right)}}\\
= \frac{{5.2.\left( {x + 3} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \frac{{10x + 30}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
+ )\frac{3}{{\left( {2x + 4} \right)\left( {x + 3} \right)}}\\
= \frac{3}{{2\left( {x + 2} \right)\left( {x + 3} \right)}}\\
= \frac{{3.3x\left( {x – 2} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \frac{{9{x^2} – 18x}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}
\end{array}$